This has nothing to do with quadratic equations. This is not algebra but calculus.
You have to take the derivative of :
Thus, following first principles,
Have you done derivatives ? Can you see how to do it now ?
(a)Find the value of m for which the line is a tangent to the curve and find the x-coordinate of the point at which this tangent touches the curve.
(b) Find the value of c and of d for which is the solution set of
For (a), how do I actually solve the quadratic equation when the equation have a ?
No idea how to solve (b)... The whole question is just like alien to me....
hi punch ,
for (1) , you multiply the whole thing by x so you don end up with
**
If a line is a tangent to the curve , it touches the curve at one point .
so
calculate the value of m , then put this value of m into ** , evaluate the x-coordinate then .
for (2) , you are given the range , so
compared this with , ?? (i think you have a typo here)
we see that c=2 , d=15 assuming that i guess it correctly .
Forgive my last post, I guess there must be a way to do it with algebra ... the hard way.
Forgive also what I said in this post, lol, I must be tired.
As mathaddict said, try to multiply the equations to get rid of the fractions and revert back to the good old quadratic equation. It makes it a lot easier.
What quadratic equation are you talking about? There is no quadratic equation here. A line crosses a graph if the equation of the line and the equation of the graph have a common factor- that is, if . The line is tangent to the curve if that is a double root. To get a quadratic equation, multiply both sides of the equation by x.
For (b), start by finding solutions to the equation . Those two points must separate "<" from ">". Try a single point, say x= 0, to determine which is which.